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A boy whirls a stone in a horizontal circle of radius 1.1 m and at height 2.1 m above ground level. The string breaks, and the stone flies horizontally and strikes the ground after traveling a horizontal distance of 10 m. What is the magnitude of the centripetal acceleration of the stone while in circular motion? m/s2

Answer :

opudodennis

Answer:

[tex]212.8 m/s^{2}[/tex]

Explanation:

Time taken by stone to cover horizontal distance

[tex]t=\sqrt{\frac {2h}{g}}[/tex] where t is time, h is height of whirling the stone in horizontal circle, g is gravitational constant, Substituting h for 2.1 m and g for 9.81

[tex]t=\sqrt{\frac {2*2.1}{9.81}}[/tex]= 0.654654 seconds

t=0.65 s

Velocity, v= distance/time

v=10/0.65= 15.27525 m/s

v=15.3 m/s

[tex]a=\frac {v^{2}}{r}[/tex] where r is radius of circle, substituting r with 1.1m

[tex]a=\frac {15.3^{2}}{1.1}[/tex]

[tex]a=212.8 m/s^{2}[/tex]

Therefore, centripetal acceleration is [tex]212.8 m/s^{2}[/tex]

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